Question: Find $\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $1$ (Choice C) C $0$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=0$ into $\dfrac{7x-\sin(x)}{x^2+\sin(3x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{7-\cos(x)}{2x+3\cos(3x)} \\\\ &=\dfrac{7-\cos(0)}{2(0)+3\cos(3\cdot0)} \gray{\text{Substitution}} \\\\ &=2 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)}=2$.